= Solution
The triangular roof has height $(w/2)\tan\theta$ and two slopes of length $(w/2)\sec\theta$. Hence the area constraint and cost are
$$
A=wh+\frac{w^2}{4}\tan\theta,
\qquad C=2\alpha h+\beta w\sec\theta.
$$
The wall height is fixed, so its cost is constant. Assume the physical case $A,h,\beta>0$. Minimize $C$ subject to the area constraint by <Lagrange multipliers>:
$$
\beta\sec\theta=\lambda\left(h+\frac w2\tan\theta\right),\qquad
\beta w\sec\theta\tan\theta=\lambda\frac{w^2}{4}\sec^2\theta.
$$
Eliminating the positive common factors gives
$$
w\sec^2\theta=4h\tan\theta+2w\tan^2\theta,
\qquad w(1-\tan^2\theta)=4h\tan\theta.
$$
Thus the optimum has $0<\theta<\pi/4$ and
$$
\boxed{2\tan2\theta=\frac wh.}
$$
Set $D=A-wh>0$, so the area constraint gives $\tan\theta=4D/w^2$. Substituting into the preceding equation and multiplying by $w^3$ yields $w^4-16D^2=16hwD$. Therefore
$$
\boxed{w^4=16D(D+wh)=16A(A-wh).}
$$
For a global check, the square of the total roof length is
$$
Q(w)=w^2+\frac{16(A-wh)^2}{w^2},\qquad0<w<A/h.
$$
Its <derivative> has the sign of $P(w)=w^4+16Ahw-16A^2$. Since $P'(w)>0$, $P(0)<0$ and $P(A/h)>0$, it has one root in this interval, with $Q$ decreasing before that root and increasing afterwards. This proves the unique global minimum, rather than merely a stationary design. The <minimum roof length at fixed wall height and cross-sectional area> is independent of $\alpha$ and of the positive factor $\beta$. If $\beta=0$, every feasible shape has equal cost; the displayed conditions then describe a possible minimizer but are not necessary.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-1-roof-optimization.png]
{title=An original cross-section construction showing the fixed wall height, roof rise and sloping lengths}
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